Solving a 2-Dimensional Collision Problem with Momentum Conservation

  • Thread starter Thread starter yaminohohenheim
  • Start date Start date
  • Tags Tags
    Collisions
Click For Summary
SUMMARY

The discussion centers on solving a 2-dimensional collision problem involving two pucks on a frictionless surface. The first puck, weighing 0.35 kg, moves at 2.3 m/s and collides with a stationary 0.23 kg puck. After the collision, the first puck moves at 2.0 m/s at an angle of -32°. The velocity of the second puck after the collision is calculated to be 1.4 m/s at an angle of 49.2° in the positive x and y quadrants. The key principle applied is the conservation of momentum, rather than kinetic energy.

PREREQUISITES
  • Understanding of 2-dimensional momentum conservation
  • Familiarity with vector decomposition in physics
  • Knowledge of basic collision theory
  • Ability to perform calculations involving angles and velocities
NEXT STEPS
  • Study the principles of momentum conservation in elastic and inelastic collisions
  • Learn about vector decomposition and how to resolve forces into components
  • Explore examples of 2-dimensional collision problems in physics textbooks
  • Practice solving collision problems using simulation tools or physics software
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding collision dynamics and momentum conservation principles in two-dimensional scenarios.

yaminohohenheim
Messages
4
Reaction score
0
can anyone offer up some help? I've been working on this forever with no success.

On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

(____m/s, _____°)


any help would be very appreciated. :-)
 
Physics news on Phys.org
Hint: What quantity is conserved during the collision?
 
Is the speed of the second puck after the collision:

v = 0,9191 i - 1,613 j

?
 
shogunultra said:
Is the speed of the second puck after the collision:

v = 0,9191 i - 1,613 j

?
I believe the question requires the answer in the form of a speed with a direction (degrees above east)
 
yaminohohenheim said:
can anyone offer up some help? I've been working on this forever with no success.

On a frictionless surface, a 0.35 kg puck moves horizontally to the right (at an angle of 0°) and a speed of 2.3 m/s. It collides with a 0.23 kg puck that is stationary. After the collision, the puck that was initially moving has a speed of 2.0 m/s and is moving at an angle of −32°. What is the velocity of the other puck after the collision?

(____m/s, _____°)


any help would be very appreciated. :-)

After my calculations, I got v = 1.4 ms-1 and in the direction of 49.2 deg in the positive x and positive y quadrant. It's better to use conservation of kinetic energy to find the resultant velocity of the stationary puck after collision. Then use conservation of total momentum in the vertical displacement to find the angle of deviation.
 
Folks, provide help, not complete solutions.

For this problem, the only thing you know is conserved is momentum. No need to assume that kinetic energy is conserved.
 

Similar threads

Replies
4
Views
2K
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
7
Views
2K
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K